Friday, January 29, 2016

Gandalf...

It has come to my attention that I am not very blog-savvy.  On 10.3 #9, I
am having trouble setting up how to do the problem.  To get from point
(3,2) to (4,7) starting with a vector 5i+2j and turning once at 90 degrees
to get to that specific endpoint.  I know that in order to be orthogonal
the dot products of the 2 vectors needs to be zero....but I don't know what
2 vectors to use and then once I have the two vectors to use....do I just
subtract or add one of those vectors to the start or end point...?  Please
help!










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OK the direction of the Iron Hills from the forest of Mirkwood is <1,5>=<4,7>-<3,2>.  Good old Gandalf doesn't go that way though (because that would be too simple).  Instead he goes in the direction of the vector <5,2> for awhile (which is *not* the same direction of <1,5>), and after awhile takes a single right turn and heads in that direction and ends up at <1,5> (I emphasize the words "right turn" because that is the crucial bit of information that tells us what we need to do)

So the picture looks like this:

If you put it in this way, it should gradually become clear that this is a problem about orthogonal projection. The right pointing red arrow is the orthogonal projection of <1,5> on <5,2>, i.e.
=(<1,5>.<5,2>)/(<5,2>.<5,2>)<5,2>=15/29<5,2>=<75/29,30/29>
This is the point where he makes his turn.
The upward pointing red arrow is the difference or residual <1,5>-<75/29,30/29>=<-46/29,115/29>.

Wednesday, January 27, 2016

Monday, January 25, 2016

Homework due this week, plus wondering...

Section 10.3

Section 10.4


Should I give a quiz on Friday? The ideas in these sections will be used all semester long...

Friday, January 22, 2016

Problem 10.2#17

 Hello Dr. Taylor,
I am can't figure out what to multiply the first part of
number 17 by to get half its length and it in the opposite direction. I got
the unit vector but can't figure out the proper scalar please help.








(click on image to enlarge)
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Well, for part 1 you're doing good things with dividing by the length to find a unit vector, but you're making it harder than it needs to be.  To get a vector half the length of u, you just have to multiply it by 1/2.  And to get it going in the opposite direction you just have to multiply it by -1. You *don't* need to divide by the length of u.

For part 2, though, you need to get a vector of exactly length 3 pointing in the direction of v, so you're doing exactly the right thing by computing the unit vector v/|v| and then multiplying by 3